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  • Evaluate   <img alt="\mathrm{\lim _{x \rightarrow 0}\lfloor\tan x\rfloor \cdot \cos x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%200

Evaluate   \mathrm{\lim _{x \rightarrow 0}\lfloor\tan x\rfloor \cdot \cos x}

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

Does not exist


Answers (1)

best_answer

\mathrm{\begin{aligned} & \lim _{x \rightarrow 0}\lfloor\tan x\rfloor \cos x= \\ & \lim _{x \rightarrow 0} \begin{cases}-\cos x & x<0 \\ 0 & x>0\end{cases} \end{aligned}}

Now if we consider the upper and lower limits :

\mathrm{\lim _{x \rightarrow 0^{-}}\lfloor\tan x\rfloor \cos x} we assume x is within \mathrm{\frac{\pi }{4}} of 0 ; that \mathrm{x\neq 0}; and we assume \mathrm{x< 0}, So,

\mathrm{\begin{aligned} & \lim _{x \rightarrow 0^{-}}\lfloor\tan x\rfloor \cos x= \\ & \lim _{x \rightarrow 0^{-}} \begin{cases}-\cos x & x<0 \\ 0 & x>0\end{cases} \\ & \lim _{x \rightarrow 0^{-}}-\cos x=-1 . \end{aligned}}

And for upper limits we assume \mathrm{x> 0} So

\mathrm{\begin{aligned} & \lim _{x \rightarrow 0^{+}}\lfloor\tan x\rfloor \cos x= \\ & \lim _{x \rightarrow 0^{+}}\left\{\begin{array}{ll} -\cos x & x<0 \\ 0 & x>0 \end{array}=\right. \\ & \lim _{x \rightarrow 0^{+}} 0=0 . \end{aligned}}

And as \mathrm{\lim _{x \rightarrow 0^{-}}\lfloor\tan x\rfloor \cos x \neq \lim _{x \rightarrow 0^{+}}\lfloor\tan x\rfloor \cos x, \lim _{x \rightarrow 0}\lfloor\tan x\rfloor \cos x}

 

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Nehul

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